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### Course: AP®︎/College Statistics>Unit 9

Lesson 4: Sampling distributions for sample proportions

# Sampling distribution of sample proportion part 1

Formulas for the mean and standard deviation of a sampling distribution of sample proportions.

## Want to join the conversation?

• This might be a silly doubt but at its taken the standard deviation for our sample proportion as the standard deviation of binomial random variable X divided by n.
Why?
• ``P-hat = X/nVar(P-hat) = Var (X/n)                         = (1/n)^2 ⋅ Var(X)                        =Var(X)/n^2                        =(n⋅p⋅(1-p))/n^2                        = p(1-p)/n  ``
• , what is the Bernoulli random variable? Was it introduced somewhere else?
• IDK if it was introduced before, but Bernoulli random variables can only take values 0 or 1. Failure or success. Yes or No. This variable is binary, because there are 2 possible results, like flipping a coin.
• could anyone explain the mean of sample proportion? (at )
• Let's say you do 99999 times of sampling tests, of sample size 10 each. You find the p̂ of each sample test, and get p̂1, p̂2 all the way to p̂99999.
p̂1 can be written as (X1)/10, p̂2 can be written as (X2)/10....p̂99999 can be written as (X99999)/10, where X1 is Total # of Yellow Gumbals in Trail No.1 & X2 is Total # of Yellow Gumballs in Trail No.2....X99999 is Total # of Yellow Gumballs in Trail No.99999.

Now, μp̂ = (p̂1 + p̂2 +...+ p̂99999) /99999. (The mean of all 99999 p̂)
We can rewrite this as [(X1)/10+(X2)/10+...(X99999)/10]/99999.
We slimplify it to [(X1+x2+...X99999)/10]/99999
= [(X1+x2+...X99999)]/[10*99999]

Since, (X1+x2+...X99999)/99999 = μX (i.e. Sum of all # of Yellow Gumballs in all 99999 trails divided by 99999, which is total number of trails).
We can finally simplify [(X1+x2+...X99999)]/[10*99999] to μX/10,
hence μp̂ = μX/n, and our example, μp̂ = μX/99999.

Sorry for the formatting, write it out on your own and it will all make sense.
• Why do we divide all the statistics (mean, SD) calculated for the sampling distribution by 'n'? I'm really confused. I mean, why do we divide the mean of the random variable X by 'n' to get the mean of the sampling distribution?
• Hi Sal...the video is very informative and well explained.

I have one question:
In the previous videos(https://www.khanacademy.org/math/statistics-probability/summarizing-quantitative-data/variance-standard-deviation-sample/v/sample-variance), you explained about the calculating the sample variance and the unbiased estimator by dividing by n-1. How that is related to this topic of sample proportion. Here also, you calculated the variance of sampling distribution of sample proportion.
Could you please explain the relation?
• I've been having a play with this same question in my head. I believe Sal takes the above standard deviation of X approach, because he explicitly states he knows the Population probability (P), 0.6. In the examples where Sal divides by n-1, we're only given sample estimates of P, P-hat, therefore we have to take into account the potential bias introduced, by using estimators instead of actual values. I think the way to mitigate this bias is by dividing by n-1 rather than n, but I might be wrong..

This video:(https://www.khanacademy.org/math/statistics-probability/sampling-distributions-library/sample-proportions/v/sampling-distribution-of-sample-proportion-part-1), uses n-1 in that context and the above is the only way I've been able to reason why!

It would be interesting to see a video where Sal had to infer the probability P, given only a sample size n and the proportion of yellow balls found x-hat to see how the approach would differ.
• Why is the square root of p*p-1 equal to the standard deviation? What makes the formula work?
• Good question!

Actually the standard deviation of Y is square root of the quantity p*(1-p).

In the lesson, Y is a random variable that is 1 with probability p, and 0 with probability (1-p).
The mean of Y, mu_Y, is E(Y) = 0*P(Y=0)+1*P(Y=1) = 0(1-p)+1*p = p.
The variance of Y, sigma^2_Y, is by definition the expected value of the squared difference of Y from its own mean.
So the variance of Y is
sigma^2_Y = E[(Y - mu_Y)^2] = E[(Y-p)^2]
= (0-p)^2 P(Y=0) + (1-p)^2 P(Y=1)
= p^2 * (1-p) + (1-p)^2 * p
= p(1-p)[p + (1-p)]
= p(1-p)(1)
= p(1-p).

The standard deviation of Y, sigma_Y, is by definition the square root of its variance, so the standard deviation of Y is sigma_Y = sqrt[p(1-p)].
• Is proportion the same as probability? The proportion is 0.6 at . So the probability of picking a yellow ball out of the machine is 60 percent. So does that mean the probability is the same as proportion? Why use a different word?
• The sample proportion is the experimental probability (based on data), but might not be the same as the theoretical (true) probability.
• At Sal says, "Well, you could say..." and the statement follows on which the rest of video relies: p-hat = X / n. While "intuitively" I find this statement valid/acceptable/../OK, I am wondering: isn't there a (more) formal argument/proof to support it?

Now that I have been thinking some more about this: isn't `p-hat`, in this context, `X/n` by definition ?